What is the value of ∆ if (cos6x + 6cos4x + 15cos2x + 10)/(cos5x + 5cos3x + 10cosx)= ∆cosx ?

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Solution

cos 6x + 6 cos 4x + 15 cos 2x + 10/ cos 5x + 5 cos 3x + 10 cos x

= cos 6x + cos 4x + 5 cos 4x + 5 cos 2x + 10 cos 2x + 10/ cos 5x + 5 cos 3 x + 10 cosx

as cos 6x + cos 4x = 2 cos 5x cosx ; cos 4 x + cos 2x = 2 cos 3 xcosx; cos 2 x + cos 0 = 2 cosxcosx

= 2 cos 5xcosx + 5 (2 cos 3 xcosx) + 10 (2 cosxcosx)/ cos 5x + 5 cos 3 x + 10 cos x

= 2 cos x (cos 5x + 5 cos 3x + 10 cosx)/ cos 5x + 5cos 3x + 10 cosx

=2cosx

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